# Finding equation of normal to a curve.

• Jul 17th 2010, 04:29 AM
jgv115
Finding equation of normal to a curve.
Well I got this question:

Differentiate from first principals (i did it) $\displaystyle f=x^3$ $\displaystyle f'(x) = 3x^2$

Find the equation of the normal to the curve at point where x = 3.

So first I found y. $\displaystyle y=(3)^3 y=27$ That means we got $\displaystyle (3,27)$

Then I subbed 3 into $\displaystyle 3x^2$ to get the gradient of the tangent. Answer is $\displaystyle 27$

Since the normal and tangent are perpendicular the formula is $\displaystyle m_{2} = \frac {-1}{m_{1}}$ So that makes $\displaystyle \frac {-1}{27}$

Then I subbed it into $\displaystyle y-y_{1} = m(x-x_{1}$

$\displaystyle y-27=\frac{-1}{27}(x+3)$

Finally I get

$\displaystyle -27y+726=x+3$ but for some reason the answers in the book says it's different :(

Did I do anything wrong?

Also, I need help with another question:

If $\displaystyle y=x^3-3x^2$ find the derivative and find the values of y when $\displaystyle \frac{dy}{dx} = 0$

How would I go about starting this one?
• Jul 17th 2010, 05:04 AM
earboth
Quote:

Originally Posted by jgv115
Well I got this question:

Differentiate from first principals (i did it) $\displaystyle f=x^3$ $\displaystyle f'(x) = 3x^2$

Find the equation of the normal to the curve at point where x = 3.

So first I found y. $\displaystyle y=(3)^3 y=27$ That means we got $\displaystyle (3,27)$

Then I subbed 3 into $\displaystyle 3x^2$ to get the gradient of the tangent. Answer is $\displaystyle 27$

Since the normal and tangent are perpendicular the formula is $\displaystyle m_{2} = \frac {-1}{m_{1}}$ So that makes $\displaystyle \frac {-1}{27}$

Then I subbed it into $\displaystyle y-y_{1} = m(x-x_{1})$

$\displaystyle y-27=\frac{-1}{27}(x-3)$ <=== you used the wrong sign in the bracket

Finally I get

$\displaystyle -27y+726=x+3$ but for some reason <=== maybe they believe that 27² = 729
the answers in the book says it's different :(

Did I do anything wrong?

Also, I need help with another question:

If $\displaystyle y=x^3-3x^2$ find the derivative and find the values of y when $\displaystyle \frac{dy}{dx} = 0$

How would I go about starting this one?

To your last question: I know that you know how to find the drivatie: Differentiate y wrt x. Solve the equation for x:

$\displaystyle 3x^2-6x = 0$
• Jul 17th 2010, 11:52 PM
jgv115
Quote:

Originally Posted by earboth
To your last question: I know that you know how to find the drivatie: Differentiate y wrt x. Solve the equation for x:

$\displaystyle 3x^2-6x = 0$

mm what do you mean i put the wrong sign? I'm pretty sure x is 3.

What does y wrt x mean?
• Jul 18th 2010, 12:11 AM
Prove It
It's actually the next step which is wrong, since when you multiply both sides by $\displaystyle -27$ you should get

$\displaystyle -27y + 729 = x - 3$.

"wrt" means "with respect to". So Earboth is saying find $\displaystyle \frac{dy}{dx}$.
• Jul 18th 2010, 12:28 AM
jgv115
Quote:

Originally Posted by Prove It
It's actually the next step which is wrong, since when you multiply both sides by $\displaystyle -27$ you should get

$\displaystyle -27y + 729 = x - 3$.

"wrt" means "with respect to". So Earboth is saying find $\displaystyle \frac{dy}{dx}$.

Yea, I actually typed it wrong.

I was meant to type:
$\displaystyle -27y + 726 = x$. I got 729 then I had to subtract 3 and i get 726. The answer says it's 732 though :S

And why do you have the equate the $\displaystyle \frac {dy}{dx}$ to 0?

$\displaystyle 3x^2-6x = 0$

$\displaystyle 3x(x-2) =0$

$\displaystyle x = 0$ $\displaystyle x= 2?$
• Jul 18th 2010, 01:10 AM
Prove It
Because that's what you were told to do in the question...

You have found the $\displaystyle x$ values of the stationary points, now substitute back into the original function to find the corresponding $\displaystyle y$ values.
• Jul 18th 2010, 01:49 AM
jgv115
y=0 and y=4

yay! :)
• Jul 18th 2010, 01:51 AM
Prove It
Quote:

Originally Posted by jgv115
Yea, I actually typed it wrong.

I was meant to type:
$\displaystyle -27y + 726 = x$. I got 729 then I had to subtract 3 and i get 726. The answer says it's 732 though :S

And why do you have the equate the $\displaystyle \frac {dy}{dx}$ to 0?

$\displaystyle 3x^2-6x = 0$

$\displaystyle 3x(x-2) =0$

$\displaystyle x = 0$ $\displaystyle x= 2?$

Like I said, you should have

$\displaystyle -27y + 729 = x - 3$.

Now you have to ADD 3 to both sides (to undo the -3 ) which gives

$\displaystyle -27y + 732 = x$ as required.
• Jul 18th 2010, 01:56 AM
jgv115
yea i get it!! (stupid mistake)